Question

Calculate the ‘spin only’ magnetic moment of ${M^{2 + }}(aq)$ ion (atomic number Z=27).

Answer 1

Magnetic moment of an atom or ion depends on the spin of its electron in an orbit. Magnetic field is produced by a moving electron in an orbit around the nucleus. In an atom, in an orbit there are maximum two electrons with opposite spin and if there is only one electron in an orbit then it creates some magnetic moment.

Complete step by step answer:
The atomic number of $M$ is 27.
In an atomic state, electronic configuration of $M$ is $\left[ {Ar} \right]3{d^7}4{s^2}$.
In ${M^{2 + }}(aq)$ ion state, it loses two electron from $4s$ orbit and its electronic configuration become $\left[ {Ar} \right]3{d^7}4{s^0}$.
Then, 3d subshell of ${M^{2 + }}(aq)$ is $\begin{gathered} \boxed{ \uparrow \downarrow }\boxed{ \uparrow \downarrow }\boxed \uparrow \boxed \uparrow \boxed \uparrow \\\ \\\ \end{gathered}$.
Here, the number of unpaired electrons is 3.
We know that, magnetic moment of an atom is given by
$\mu = \sqrt {n(n + 1)}$, where $n$ is the number of unpaired electrons.
$\mu = \sqrt {3(3 + 1)} = \sqrt {12} = 3.87BM$
Hence, the magnetic moment of ${M^{2 + }}(aq)$ is $3.87BM$.

Note: Magnetic moment of atom depends on number of paired electrons. If there are unpaired electron in an atom then that atom is paramagnetic and if there is no unpaired electron in atom then it is diamagnetic. Magnet attracts paramagnetic metal with a strong magnetic force but diamagnetic metal or objects are not affected by a magnet there is only a small effect on diamagnetic atoms.